Solving simple differential equation (t+4)dx=4(1+x2)dt

killjoy1990xb9

killjoy1990xb9

Answered

2022-01-20

Solving simple differential equation
(t+4)dx=4(1+x2)dt

Answer & Explanation

Bob Huerta

Bob Huerta

Expert

2022-01-20Added 41 answers

As you rearranged it in your post, we have
dx1+x2=4dtt+4
As you discovered, the LHS is arctan. To find the value of the RHS, we may say t+4=u so du=dt. Thus the RHS can be rewritten as
4duu=4log|u|+C2
Because we want this integral to be in respect to t, we have u=t+4, so
arctanx+C1=4log|t+4|+C2
To confirm that the integral we solved is indeed correct, you can simply differentiate 4log|t+4|+C2
Thus x=tan(4log|t+4|+C2C1)
and t=exp(14(arctanx+C1C2))4
Dawn Neal

Dawn Neal

Expert

2022-01-21Added 35 answers

Here are some proceeding steps:
- Seperating the variables you get
dx1+x2=4dtt+4
- Integrating both sides you have:
tan1(x)+C1=4log|t+4|+C2.

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